Question

outside diameter of 3.750 in. and a wall thickness of 0.082 in. is subjected to loads P 1400 lb, P = 970 lb, and T 8400 lb in. as shown in the figure. Using a = 6.25 in, determine the normal and A circular tube witha shear stresses that exist at (a) point H and (b) point K (a) Enter the normal and shear stresses and point H. Your answers must be consistent with the sign conventions for stresses. nsi Dы (b) Enter the normal and shear stresses and point K. Your answers must be consistent with the sign conventions for stresses psi psi pSi

A circular tube with an outside diameter of 3.750 in. and a wall thickness of 0.082 in. is subjected to loads Px=1400 lb, Pz=970 lb, and T=8400 lb⋅in. as shown in the figure. Using a=6.25 in., determine the normal and shear stresses that exist at (a) point H and (b) point K.

Enter the normal and shear stresses and point H. Your answers must be consistent with the sign conventions for stresses.

Enter the normal and shear stresses and point K. Your answers must be consistent with the sign conventions for stresses.

Please do these numbers not a picture of another question.

Answer 1

STRESSES ATH GIVEN DATA outer Diameter d_0 = 3 750 in wall thickness t = 0 082 in Normal force p_y = 0 shear force p_x = 1400 lb shear force p_z = 970 lb Bending Moments M_x = 970 times 6 25 = 6062 5 lb in M_z = 1400 times 6 25 = 8750 lb in M_z = 8400 lb in calculations inner Diameter d_i = 3 750 - 2 times 0 082 = 3 586 in cross sectional area A = pi/4 (d^2 - d^2) = 0 945 in^2 Moment of inertia 1_times = 1_2 = pi/64 (d_0^4 - d^4) = 1 590 in polar moment of inertia l)y = pi/32 (d_0^5 - d^4) = 3 180 in^4 stresses at point H normal stress sigma = 0/0 945 = 0 psi Bending stresses due to mx sigma = (6062 5) x (3 750/2)/1 590 = 7149 psi (compressive) due to sigma = 0 { as the point lies at the neutral axis)